consecutive square numbers
Multiply each preceding term by 4.
Everytime you move to the next number you add 1, then 2, then 3, etc.1+1=22+2=44+3=77+4=1111+5=1616+6=2222+7=2929+8=37
The rule is 5, 10, 15 and so the next number will be 20+31 = 51
So you're adding 3, 5, 7, 9, 11 etc 1, 4, 9, 16, 25, 36, 49
There are actually more than a pattern rule for that type of sequence. The pattern rule is not obvious. The sequence is actually this: an = 5a(n - 2) for n > 2, such that the seed values are a1 = 1 and a2 = 5. Another pattern rule is the number of palindromes using a maximum of five different symbols. The third one is the Inverse Binomial Mean transformation of the Fibonacci sequence.
t(n) = n2
Double the previous number
add 3,5,7,9and then to 25 add 11, adding odd numbers onto the last . 1+3=4, 4+5=9, 9+7=16, 16+9=25, 25+11=36 and so on
Yes
Yes.
Multiply each preceding term by 4.
t(n) = (n+1)2 where n = 1, 2, 3, ...
Everytime you move to the next number you add 1, then 2, then 3, etc.1+1=22+2=44+3=77+4=1111+5=1616+6=2222+7=2929+8=37
The rule is 5, 10, 15 and so the next number will be 20+31 = 51
assuming you mis-typed the 15 and instead it should be a 25, these are simply the squares of the number's position in the line. 12 = 1 22 = 2 32 = 9 42 = 16 52 = 25 (assume is what you meant) 62 = 36 You can also think of it as adding on the next in a series of odd numbers 1+3 = 4 4+5 = 9 9+7 = 16 which results from expanding (x+1)2 - x2 anyway
no it is not a recursive pattern because it isn't equal numbers.
from 12 to 16 is 4 then from 16 to fiftine is one then from15-19 is 4 so one so forth one is +4 the next is -1